Question

A ship's whistle, 0.600 m long, is an air column closed at one end. If the ship's speed of sound in air is 340.0 m/s, calculate the three lowest resonant frequencies for the ship's whistle.

Answer 1

We are given a closed air column and we are asked to determine the resonant frequencies. To determine the first frequency we will use the following formula:

$f=\frac{V}{4L}$

Where:

$\begin{gathered} V=\text{ sp}eed\text{ of sound} \\ L=\text{ length} \end{gathered}$

Substituting we get:

$f=\frac{(340\frac{m}{s})}{4(0.6m)}$

Solving the operations we get:

$f=142Hz$

Therefore, the first low frequency is 142 Hertz.

Now, to determine the second frequency we use the fact that a closed air column will produce odd harmonics only, therefore, the second frequency is determined by multiplying the first frequency by 3, like this:

$f_2=3f_1$

Substituting we get:

$f_2=3(142Hz)=426Hz$

Therefore, the second frequency is 426 Hz. Now, the third frequency is determined by multiplying the first frequency by 5:

$f_3=5f_1$

Substituting we get:

$f_3=5(142Hz)=710Hz$

Therefore, the third frequency is 710 Hertz.